# Math Help - Trigonometric Integrals

1. ## Trigonometric Integrals

I'm having trouble finding out how to solve this integral.

I know the correct answer is: 0.235619, but I want to know if my work is correct. So far, whenever I work it out, I don't get the right answer.

From the integral,

((1/2)(1-cos10x))^2

= (1/4) integral (1-cos10x)^2

=(1/4) integral (1-2cos10x * (cos10x)^2

=(1/4) integral 1-2cos10x * ((1/2)1+cos20x)

From there, I plugged in the numbers, but I'm not getting the correct answer. Help?

2. Originally Posted by Kimmy2
I'm having trouble finding out how to solve this integral.

I know the correct answer is: 0.235619, but I want to know if my work is correct. So far, whenever I work it out, I don't get the right answer.

From the integral,

((1/2)(1-cos10x))^2

= (1/4) integral (1-cos10x)^2

=(1/4) integral (1-2cos10x * (cos10x)^2

=(1/4) integral 1-2cos10x * ((1/2)1+cos20x)

From there, I plugged in the numbers, but I'm not getting the correct answer. Help?
$\sin ^4 5x = \frac{(1-\cos 10x)^2}{4} = \frac{1-2\cos 10x +\cos ^2 10x}{4} = \frac{1-2\cos 10x + \frac{\cos 20x +1}{2} }{4}$

$\sin ^4 5x = \frac{2 - 4\cos 10x + \cos 20x +1 }{8}$

$\sin ^4 5x = \frac{3 -4\cos 10x + \cos 20x }{8}$

when you integrate it you will get

$\frac{3\pi}{5\cdot 8}= \frac{3\pi}{40}$

Thanks!

4. Originally Posted by Kimmy2
$\int _{0}^{\frac{\pi}{5}} \sin ^4 5x \cdot dx$
$\int_{0}^{\frac{\pi}{5}} \frac{3-4\cos 10x + \cos 20x }{8} \cdot dx$
$\left(\frac{1}{8}\right) \left( 3x - \frac{4\sin 10x}{10} + \frac{\sin 20x}{20} \right)\mid^{\frac{\pi}{5}}_{0}$
$\left(\frac{1}{8}\right) \left( \frac{3\pi}{5} - \frac{4\sin 2\pi}{10} + \frac{\sin 4\pi}{20} \right)-\left(\frac{1}{8}\right) \left( 3(0) - \frac{4\sin (0)}{10} + \frac{\sin (0)}{20} \right)$
$\left(\frac{1}{8}\right) \left( \frac{3\pi}{5}\right)$